A= \(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-2\right)}:\dfrac{6}{x+2}\)
\(=\dfrac{-1}{x-2}\)
b) \(\left|x\right|=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2};x=\dfrac{-1}{2}\)
Thay \(x=\dfrac{1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne2;x\ne-2\) )
\(A=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{2}{3}\)
Thay \(x=\dfrac{-1}{2}\) (Thỏa mãn ĐKXĐ \(x\ne2;x\ne-2\) )
\(A=\dfrac{-1}{\dfrac{-1}{2}-2}=\dfrac{2}{5}\)
c) \(\dfrac{-1}{x-2}< 0\)
\(\Rightarrow x-2>0\)
\(\Leftrightarrow x>2\)
Vậy x > 2 để A < 0